Results 1 to 5 of 5

Math Help - horizontal circular motion

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    239

    horizontal circular motion

    a car of mass 800 kg is moving in a horizontal circle of radius 50m on a road banked at an angle of 8 degrees.

    When the car is moving at a constant speed of v m/s, there is no sideways force acting on the car. Calculate, to two significant figures, the normal reaction of the road on the car and the value of v.


    Right so im finding this tough because of the fact the road is banked. Usually the questions involve a piece of string and a conical pendulum. Those questions i can just about manage! So going off what i know, i'd assume is resolve the forces towards the centre - which i think is the bottom of the slope?? now normally i would also resolve the tension in the string but obviously thats not the case in this question. I'm really stumped
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    This works somewhat like the problem of a car on an incline plane

    If the track were flat then the normal force would simply be the weight.

    But because of the 8 degree bank Ncos(t) = weight = 800g

    N = 800*g/cos(8)


    The Horizontal component N sin(t) provides the centripetal force

    Nsin(8) = mv^2/R

    Using your result from the first part v^2 = Nsin(8)R/m
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2008
    Posts
    239
    This works somewhat like the problem of a car on an incline plane

    If the track were flat then the normal force would simply be the weight.

    But because of the 8 degree bank Ncos(t) = weight = 800g

    N = 800*g/cos(8)
    i'm having a little trouble understanding this. does N denote the reaction of the car on the slope? can this question be interpreted as a conical pendulum question? the method seems very similar

    The Horizontal component N sin(t) provides the centripetal force

    Nsin(8) = mv^2/R

    Using your result from the first part v^2 = Nsin(8)R/m
    i understand this thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    There are 3 forces here

    N is the force between the car and the road.

    It has a vertical and horizontal component

    the vertical component Ncos(t) is just the weight

    the horizontal component is the centripetal force

    It is similar to the conical pendulum problem in that the tension T is replaced by the normal force N

    In fact for the conical pendulum Tcos(t) = w

    Tsin(t) = mv^2/R

    exactly the same as for the car where the angle of the string is replaced by the banking angle of the track.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2008
    Posts
    239
    In fact for the conical pendulum Tcos(t) = w

    Tsin(t) = mv^2/R

    exactly the same as for the car where the angle of the string is replaced by the banking angle of the track.
    this is what i thought Much appreciated!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Circular Motion
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: October 20th 2010, 10:25 AM
  2. Circular Motion.
    Posted in the Advanced Applied Math Forum
    Replies: 11
    Last Post: June 24th 2010, 05:58 AM
  3. Circular motion?
    Posted in the Advanced Applied Math Forum
    Replies: 5
    Last Post: December 28th 2008, 09:52 PM
  4. Circular Motion
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 4th 2006, 05:56 AM

Search Tags


/mathhelpforum @mathhelpforum